Question

Find all solutions in the interval [0,2л).

sin 2x cos x + sinx = 0

x =

Answer 1

`x=0; x= \pi`

Explanation:

Let's first convert that
`sin\ 2x = 2 sinx \ cosx` using the double angle formula.

`2 sinx\ cos^2x + sinx = 0`

Now, if
`sin x = 0 \implies x= k\pi` with
`k\in \{0, 1\}` we get
`0=0` which is indeed true, so we have a solution. Else, we can divide by
`sin x \\e 0` to get
`cos^2x +1 = 0` which has no real solutions (I like to think about it as the sum of two positive numbers, which is never zero).